Answer:
The required wall thickness is [tex]1.506 \times 10^{-3}[/tex] m
Explanation:
Given:
Fluid density [tex]\rho = 1200[/tex] [tex]\frac{kg}{m^{3} }[/tex]
Diameter of tank [tex]d = 8[/tex] m
Length of tank [tex]l = 4[/tex] m
F.S = 4
For A-36 steel yield stress [tex]\sigma = 250[/tex] MPa,
Allowable stress [tex]\sigma _{allow} = \frac{\sigma}{F.S}[/tex]
[tex]\sigma _{allow} = \frac{250}{4} = 62.5[/tex] MPa
Pressure force is given by,
[tex]P = \rho gh[/tex]
[tex]P = 1200 \times 9.8 \times 4[/tex]
[tex]P = 47088[/tex] Pa
Now for a vertical pipe,
[tex]\sigma _{allow} = \frac{Pd}{4t}[/tex]
Where [tex]t =[/tex] required thickness
[tex]t = \frac{Pd}{4 \sigma _{allow} }[/tex]
[tex]t = \frac{47088 \times 8 }{4 \times 62.5 \times 10^{6} }[/tex]
[tex]t = 1.506 \times 10^{-3}[/tex] m
Therefore, the required wall thickness is [tex]1.506 \times 10^{-3}[/tex] m